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反比例函数y=kx(k>1)和y=1x在第一象限内的图象如图所示,点P在y=kx的图象上,PC⊥x轴于C,交y=1x的图象于A,PD⊥y轴于D,交y=1x的图象于B,当点P在反比例函数y=kx上运动时,以下结论①S△ODB=

题目详情
反比例函数y=
k
x
(k>1)和y=
1
x
在第一象限内的图象如图所示,点P在y=
k
x
的图象上,PC⊥x轴于C,交y=
1
x
的图象于A,PD⊥y轴于D,交y=
1
x
的图象于B,当点P在反比例函数y=
k
x
上运动时,以下结论①S△ODB=
1
2
;②四边形PAOB的面积始终不变;③PA=PB;④
PD
PB
=
PC
PA
;其中一定正确的是(  )

A.①②③
B.①③④
C.①②④
D.①②
k
x
(k>1)和y=
1
x
在第一象限内的图象如图所示,点P在y=
k
x
的图象上,PC⊥x轴于C,交y=
1
x
的图象于A,PD⊥y轴于D,交y=
1
x
的图象于B,当点P在反比例函数y=
k
x
上运动时,以下结论①S△ODB=
1
2
;②四边形PAOB的面积始终不变;③PA=PB;④
PD
PB
=
PC
PA
;其中一定正确的是(  )

A.①②③
B.①③④
C.①②④
D.①②
k
x
kkxx
1
x
在第一象限内的图象如图所示,点P在y=
k
x
的图象上,PC⊥x轴于C,交y=
1
x
的图象于A,PD⊥y轴于D,交y=
1
x
的图象于B,当点P在反比例函数y=
k
x
上运动时,以下结论①S△ODB=
1
2
;②四边形PAOB的面积始终不变;③PA=PB;④
PD
PB
=
PC
PA
;其中一定正确的是(  )

A.①②③
B.①③④
C.①②④
D.①②
1
x
11xx
k
x
的图象上,PC⊥x轴于C,交y=
1
x
的图象于A,PD⊥y轴于D,交y=
1
x
的图象于B,当点P在反比例函数y=
k
x
上运动时,以下结论①S△ODB=
1
2
;②四边形PAOB的面积始终不变;③PA=PB;④
PD
PB
=
PC
PA
;其中一定正确的是(  )

A.①②③
B.①③④
C.①②④
D.①②
k
x
kkxx
1
x
的图象于A,PD⊥y轴于D,交y=
1
x
的图象于B,当点P在反比例函数y=
k
x
上运动时,以下结论①S△ODB=
1
2
;②四边形PAOB的面积始终不变;③PA=PB;④
PD
PB
=
PC
PA
;其中一定正确的是(  )

A.①②③
B.①③④
C.①②④
D.①②
1
x
11xx
1
x
的图象于B,当点P在反比例函数y=
k
x
上运动时,以下结论①S△ODB=
1
2
;②四边形PAOB的面积始终不变;③PA=PB;④
PD
PB
=
PC
PA
;其中一定正确的是(  )

A.①②③
B.①③④
C.①②④
D.①②
1
x
11xx
k
x
上运动时,以下结论①S△ODB=
1
2
;②四边形PAOB的面积始终不变;③PA=PB;④
PD
PB
=
PC
PA
;其中一定正确的是(  )

A.①②③
B.①③④
C.①②④
D.①②
k
x
kkxx△ODB
1
2
;②四边形PAOB的面积始终不变;③PA=PB;④
PD
PB
=
PC
PA
;其中一定正确的是(  )

A.①②③
B.①③④
C.①②④
D.①②
1
2
1122
PD
PB
=
PC
PA
;其中一定正确的是(  )

A.①②③
B.①③④
C.①②④
D.①②
PD
PB
PDPDPBPB
PC
PA
;其中一定正确的是(  )

A.①②③
B.①③④
C.①②④
D.①②
PC
PA
PCPCPAPA




▼优质解答
答案和解析
设P(m,n),则mn=k,
∵A、B两点在双曲线y=
1
x
上,
∴A(m,
1
m
),B(
1
n
,n),
∴①S△ODB=
1
2
DB×OD=
1
2
×
1
n
×n=
1
2
,结论正确;
②S四边形PAOB=S矩形OCPD-S△OBD-S△OAC=mn-
1
2
-
1
2
=k-1(定值),结论正确;
③PA=n-
1
m
=
k−1
m
,PB=m-
1
n
=
k−1
n
,PA≠PB,结论错误;
PD
PB
=
m
m−
1
n
=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
1
x
111xxx上,
∴A(m,
1
m
),B(
1
n
,n),
∴①S△ODB=
1
2
DB×OD=
1
2
×
1
n
×n=
1
2
,结论正确;
②S四边形PAOB=S矩形OCPD-S△OBD-S△OAC=mn-
1
2
-
1
2
=k-1(定值),结论正确;
③PA=n-
1
m
=
k−1
m
,PB=m-
1
n
=
k−1
n
,PA≠PB,结论错误;
PD
PB
=
m
m−
1
n
=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
1
m
111mmm),B(
1
n
,n),
∴①S△ODB=
1
2
DB×OD=
1
2
×
1
n
×n=
1
2
,结论正确;
②S四边形PAOB=S矩形OCPD-S△OBD-S△OAC=mn-
1
2
-
1
2
=k-1(定值),结论正确;
③PA=n-
1
m
=
k−1
m
,PB=m-
1
n
=
k−1
n
,PA≠PB,结论错误;
PD
PB
=
m
m−
1
n
=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
1
n
111nnn,n),
∴①S△ODB△ODB=
1
2
DB×OD=
1
2
×
1
n
×n=
1
2
,结论正确;
②S四边形PAOB=S矩形OCPD-S△OBD-S△OAC=mn-
1
2
-
1
2
=k-1(定值),结论正确;
③PA=n-
1
m
=
k−1
m
,PB=m-
1
n
=
k−1
n
,PA≠PB,结论错误;
PD
PB
=
m
m−
1
n
=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
1
2
111222DB×OD=
1
2
×
1
n
×n=
1
2
,结论正确;
②S四边形PAOB=S矩形OCPD-S△OBD-S△OAC=mn-
1
2
-
1
2
=k-1(定值),结论正确;
③PA=n-
1
m
=
k−1
m
,PB=m-
1
n
=
k−1
n
,PA≠PB,结论错误;
PD
PB
=
m
m−
1
n
=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
1
2
111222×
1
n
×n=
1
2
,结论正确;
②S四边形PAOB=S矩形OCPD-S△OBD-S△OAC=mn-
1
2
-
1
2
=k-1(定值),结论正确;
③PA=n-
1
m
=
k−1
m
,PB=m-
1
n
=
k−1
n
,PA≠PB,结论错误;
PD
PB
=
m
m−
1
n
=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
1
n
111nnn×n=
1
2
,结论正确;
②S四边形PAOB=S矩形OCPD-S△OBD-S△OAC=mn-
1
2
-
1
2
=k-1(定值),结论正确;
③PA=n-
1
m
=
k−1
m
,PB=m-
1
n
=
k−1
n
,PA≠PB,结论错误;
PD
PB
=
m
m−
1
n
=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
1
2
111222,结论正确;
②S四边形PAOB四边形PAOB=S矩形OCPD矩形OCPD-S△OBD△OBD-S△OAC△OAC=mn-
1
2
-
1
2
=k-1(定值),结论正确;
③PA=n-
1
m
=
k−1
m
,PB=m-
1
n
=
k−1
n
,PA≠PB,结论错误;
PD
PB
=
m
m−
1
n
=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
1
2
111222-
1
2
=k-1(定值),结论正确;
③PA=n-
1
m
=
k−1
m
,PB=m-
1
n
=
k−1
n
,PA≠PB,结论错误;
PD
PB
=
m
m−
1
n
=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
1
2
111222=k-1(定值),结论正确;
③PA=n-
1
m
=
k−1
m
,PB=m-
1
n
=
k−1
n
,PA≠PB,结论错误;
PD
PB
=
m
m−
1
n
=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
1
m
111mmm=
k−1
m
,PB=m-
1
n
=
k−1
n
,PA≠PB,结论错误;
PD
PB
=
m
m−
1
n
=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
k−1
m
k−1k−1k−1mmm,PB=m-
1
n
=
k−1
n
,PA≠PB,结论错误;
PD
PB
=
m
m−
1
n
=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
1
n
111nnn=
k−1
n
,PA≠PB,结论错误;
PD
PB
=
m
m−
1
n
=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
k−1
n
k−1k−1k−1nnn,PA≠PB,结论错误;
PD
PB
=
m
m−
1
n
=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
PD
PB
PDPDPDPBPBPB=
m
m−
1
n
=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
m
m−
1
n
mmmm−
1
n
m−
1
n
m−
1
n
111nnn=
k
k−1
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
k
k−1
kkkk−1k−1k−1,
PC
PA
=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
PC
PA
PCPCPCPAPAPA=
n
n−
1
m
=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
n
n−
1
m
nnnn−
1
m
n−
1
m
n−
1
m
111mmm=
k
k−1
PD
PB
=
PC
PA
,结论正确.
故选C.
k
k−1
kkkk−1k−1k−1,
PD
PB
=
PC
PA
,结论正确.
故选C.
PD
PB
PDPDPDPBPBPB=
PC
PA
,结论正确.
故选C.
PC
PA
PCPCPCPAPAPA,结论正确.
故选C.